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If the pressure of gas is increased from \(1.01 \times 10^{5} \mathrm{~Pa}\) to \(1.165 \times 10^{5} \mathrm{~Pa}\) and volume is decreased by \(10 \%\) at constant temperature, then the bulk modulus of the gas is (a) \(15.5 \times 10^{5} \mathrm{~Pa}\) (b) \(1.4 \times 10^{5} \mathrm{~Pa}\) (c) \(1.55 \times 10^{5} \mathrm{~Pa}\) (d) \(0.0155 \times 10^{5} \mathrm{~Pa}\)

Short Answer

Expert verified
The bulk modulus of the gas is \(1.55 \times 10^{5} \text{ Pa}\), which corresponds to option (c).

Step by step solution

01

Understanding the Bulk Modulus

The bulk modulus, denoted as \( B \), is a measure of a substance's resistance to uniform compression. It is defined as the ratio of the infinitesimal change in pressure \( \Delta P \) to the fractional change in volume \( \frac{\Delta V}{V} \) when the temperature remains constant. The formula is \( B = - \frac{\Delta P}{\frac{\Delta V}{V}} \).
02

Identifying Given Values

From the problem, we know the initial pressure \( P_1 = 1.01 \times 10^{5} \text{ Pa} \) and the final pressure \( P_2 = 1.165 \times 10^{5} \text{ Pa} \). The volume decreases by \( 10\% \), so \( \frac{\Delta V}{V} = -0.10 \).
03

Calculating Change in Pressure

Calculate the change in pressure \( \Delta P \). \[ \Delta P = P_2 - P_1 = 1.165 \times 10^{5} \text{ Pa} - 1.01 \times 10^{5} \text{ Pa} = 0.155 \times 10^{5} \text{ Pa} \]
04

Applying the Bulk Modulus Formula

Substitute \( \Delta P = 0.155 \times 10^{5} \text{ Pa} \) and \( \frac{\Delta V}{V} = -0.10 \) into the bulk modulus formula. \[ B = - \frac{0.155 \times 10^{5} \text{ Pa}}{-0.10} = 1.55 \times 10^{5} \text{ Pa} \]
05

Finding the Correct Option

Compare the calculated bulk modulus \( 1.55 \times 10^{5} \text{ Pa} \) with the provided options. The correct answer is option (c) \( 1.55 \times 10^{5} \text{ Pa} \).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Pressure Change
Pressure change refers to the difference in pressure between two states of a gas in a closed system. In this exercise, we observe an increase in the pressure of the gas from the initial pressure of \(1.01 \times 10^{5} \text{ Pa}\) to a final pressure of \(1.165 \times 10^{5} \text{ Pa}\). The change in pressure, denoted as \(\Delta P\), is the key value we need to determine when calculating the bulk modulus.Calculating the change in pressure is straightforward:
  • \(\Delta P = P_2 - P_1\)
  • Where \(P_2\) is the final pressure and \(P_1\) is the initial pressure.
In this scenario, \(\Delta P = 1.165 \times 10^{5} \text{ Pa} - 1.01 \times 10^{5} \text{ Pa} = 0.155 \times 10^{5} \text{ Pa}\).Understanding this concept is crucial when dealing with real-world applications, such as calculating the strength of materials or systems subjected to varying pressures.
Volume Change
Volume change is represented by \(\frac{\Delta V}{V}\), which describes the fractional change in the volume of a gas. It’s a key concept in determining the bulk modulus, as the formula for bulk modulus is dependent on this volume change.In the problem provided, the volume of the gas decreases by 10% when the pressure is increased. The fractional volume change can therefore be expressed mathematically as:
  • \(\frac{\Delta V}{V} = -0.10\)
The negative sign indicates that the volume is decreasing, which is typical in compression scenarios. Recognizing the symbol \(\Delta\) is crucial, as it represents a change in a quantity.Understanding volume change allows students to grasp how materials behave under pressure, particularly in the context of compression, an essential concept in engineering and material science.
Thermal Physics
Thermal physics involves the study of energy, heat, and work within physical systems. In this exercise, understanding thermal physics helps us analyze how gas behaves when pressure and volume change at a constant temperature. When dealing with gases, you often want to know:
  • How energy is transferred in the form of heat.
  • How this energy impacts the system's volume, pressure, and temperature.
In our context, the system is isothermal, meaning the temperature does not change even as pressure and volume do. Knowing the laws and principles under thermal physics is fundamental for solving problems in fluid dynamics, as it helps explain the behavior of gases during compression or expansion.
Isothermal Process
An isothermal process is a thermodynamic process in which the temperature remains constant. This is significant in ideal gas scenarios, such as in the provided exercise, where the process of compression happens without a change in temperature.Key characteristics of an isothermal process include:
  • Heat exchange with the environment to maintain constant temperature.
  • For ideal gases, the relation \(PV = nRT\) holds true, where pressure \(P\) and volume \(V\) change but \(T\) remains constant.
The isothermal condition in this exercise simplifies calculations of bulk modulus, since there's no need to account for temperature changes.Understanding isothermal processes is crucial for various scientific disciplines, particularly in determining thermal efficiencies and designing engines or refrigerators.

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Most popular questions from this chapter

A steel wire is suspended vertically from a rigid support. When loaded with a weight in air, it extends by \(l_{a}\) and when the weight is immersed completely in water, the extension is reduced to \(l_{w}\). Then the relative density of the material of the weight is (a) \(\frac{l_{a}}{l_{w}}\) (b) \(\frac{l_{a}}{l_{a}-l_{w}}\) (c) \(\frac{l_{w}}{l_{a}-l_{w}}\) (d) \(\frac{l_{m}}{l_{a}}\)

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